What is the extraneous solution to these equations? $\dfrac{x^2 - 8}{x + 9} = \dfrac{-8x + 1}{x + 9}$
Explanation: Multiply both sides by $x + 9$ $ \dfrac{x^2 - 8}{x + 9} (x + 9) = \dfrac{-8x + 1}{x + 9} (x + 9)$ $ x^2 - 8 = -8x + 1$ Subtract $-8x + 1$ from both sides: $ x^2 - 8 - (-8x + 1) = -8x + 1 - (-8x + 1)$ $ x^2 - 8 + 8x - 1 = 0$ $ x^2 - 9 + 8x = 0$ Factor the expression: $ (x + 9)(x - 1) = 0$ Therefore $x = -9$ or $x = 1$ At $x = -9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -9$, it is an extraneous solution.